Question

An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 5.4 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 24 engines and the mean pressure was 5.7 pounds/square inch with a standard deviation of 1.0. A level of significance of 0.05 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer 1

The p-value from the hypothesis test is 0.142 i.e., greater than the given significance level of 0.05. So, the null hypothesis is not rejected. The z-score for the given sample is 1.471.

What is the decision rule for the p-value approach to hypothesis testing?

The decision rule based on p-value states,

• If p > α (significance level), then the null hypothesis is not rejected
• If p < α (significance level), then the null hypothesis is rejected in favor of the alternative hypothesis.

Calculation:

Since it is given that the valve would produce a mean pressure of 5.4 pounds/square inch. I.e., μ = 5.4 p/si

So, Defining the hypothesis:

Null hypothesis H0: μ = 5.4

Alternative hypothesis Ha: μ ≠ 5.4

It is given that,

The valve was tested on 24 engines. I.e., Sample size n = 24

The sample mean X = 5.7

Standard deviation σ = 1.0 and

The significance level = 0.05

Since the population distribution is approximately normal,

the test statistic is calculated as follows:

z = (X - μ)/(σ/$\sqrt{n}$)

On substituting the value,

z = (5.7 - 5.4)/(1.0/$\sqrt{24}$)

  = (0.3)/0.204

  = 1.471

Fron this z-score, the p-value is calculated as 0.142.

Since, the value of p > 0.05 (significance level), the null hypothesis is not rejected.

Learn more about hypothesis testing here:

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